I've discovered what I believe is a quite remarkable sequence, defined by $$n_1 = 3,\qquad n_{k+1} = 2^{n_k-1}+5\quad(k\geq 1).$$ Here are the first four terms with their prime factorizations: $$ \begin{split} n_1 &= 3,\\ n_2 &= 2^2 + 5 = 9 = 3^2,\\ n_3 &= 2^8 + 5 = 261 = 3^2\cdot 29, \\ n_4 &= 2^{260} + 5 = 3^2\cdot 29\cdot p_{76}, \end{split} $$ where $p_{76}$ is a 76-digit prime.

I've tested that $n_5$ has no prime factors below $10^{12}$ besides those already present in $n_4$ (i.e., primes $3$ and $29$). Furthermore, neither of $n_6,n_7,\dots,n_{100}$ has prime factors below $10^8$, besides the same $3$ and $29$. I'm currently pushing these bounds forward, but I already wonder why the terms so massively miss small prime factors. Is there an apparent reason for such behavior?

I'm still puzzled by so massive absence of small prime factors, and plan to extend my search for prime factors to $10^{14}$ or so.

I've discovered what I believe is a quite remarkable sequence (A318970), defined by $$n_1 = 3,\qquad n_{k+1} = 2^{n_k-1}+5\quad(k\geq 1).$$ Here are the first four terms with their prime factorizations: $$ \begin{split} n_1 &= 3,\\ n_2 &= 2^2 + 5 = 9 = 3^2,\\ n_3 &= 2^8 + 5 = 261 = 3^2\cdot 29, \\ n_4 &= 2^{260} + 5 = 3^2\cdot 29\cdot p_{76}, \end{split} $$ where $p_{76}$ is a 76-digit prime.

UPDATE #4 (Oct 25 2018). I've extended the search for prime factors to $10^{14}$ and found only one more prime 28480625878963, which divides $n_k$ for all $k\geq 11$. The known prime factors are now listed in the sequence A318971.

I've tested that $n_5$ has no prime factors below $10^{12}$ besides those already present in $n_4$ (i.e., primes $3$ and $29$). Furthermore, neither of $n_6,n_7,\dots,n_{100}$ has prime factors below $10^8$, besides the same $3$ and $29$. I'm currently pushing these bounds forward, but I already wonder why the terms so massively miss small prime factors. Is there an apparent reason for such behavior?

I've tested that $n_5$ has no prime factors below $10^{12}$ besides those already present in $n_4$ (i.e., primes $3$ and $29$). Furthermore, neither of $n_6,n_7,\dots,n_{100}$ has prime factors below $10^8$, besides the same $3$ and $29$. I'm currently pushing these bounds forward, but I already wonder why the terms so massively miss small prime factors. Is there an apparent reason for such behavior?

UPDATE #3 (Jan 24 2018). As explained in UPDATE #1 above, for every prime $p$, one can compute the set of indices $k$ such that $p\mid n_k$. This set is either finite (possibly empty), or includes all large enough integers.

Recently I've tested all primes below $10^{13}$ and found that the only primes in this range that ever divide $n_k$ are $3$, $29$, and $31821709567$. The last prime divides $n_k$ for all $k\geq 8$, and so it does not help to resolve the question in favor of either outcome.

I'm still puzzled by so massive absence of small prime factors, and plan to extend my search for prime factors to $10^{14}$ or so.